Thus far we have focused mostly on 2-dimensional vector fields, measuring flow and flux along/across curves in the plane. Both Green’s Theorem and the Divergence Theorem make connections between planar regions and their boundaries. We now move our attention to 3-dimensional vector fields, considering both curves and surfaces in space.
We are accustomed to describing surfaces as functions of two variables, usually written as . For our coming needs, this method of describing surfaces will prove to be insufficient. Instead, we will parametrize our surfaces, describing them as the set of terminal points of some vector-valued function . The bulk of this section is spent practicing the skill of describing a surface using a vector-valued function. Once this skill is developed, we’ll show how to find the surface area of a parametrically-defined surface , a skill needed in the remaining sections of this chapter.
Let be a vector-valued function that is continuous and one to one on the interior of its domain in the - plane. The set of all terminal points of (i.e., the range of ) is the surface , and along with its domain form a parametrization of .
Given a point in the domain of a vector-valued function , the vectors and are tangent to the surface at (a proof of this is developed later in this section). The definition of smoothness dictates that ; this ensures that neither nor are , nor are they ever parallel. Therefore smoothness guarantees that and determine a plane that is tangent to .
A surface is said to be orientable if a field of normal vectors can be defined on that vary continuously along . This definition may be hard to understand; it may help to know that orientable surfaces are often called “two sided.” A sphere is an orientable surface, and one can easily envision an “inside” and “outside” of the sphere. A paraboloid is orientable, where again one can generally envision “inside” and “outside” sides (or “top” and “bottom” sides) to this surface. Just about every surface that one can imagine is orientable, and we’ll assume all surfaces we deal with in this text are orientable.
It is enlightening to examine a classic non-orientable surface: the Möbius band, shown in Figure 15.5.3. Vectors normal to the surface are given, starting at the point indicated in the figure. These normal vectors “vary continuously” as they move along the surface. Letting each vector indicate the “top” side of the band, we can easily see near any vector which side is the “top”.
However, if as we progress along the band, we recognize that we are labeling “both sides” of the band as the top; in fact, there are not two “sides” to this band, but one. The Möbius band is a non-orientable surface.
There is a straightforward way to parametrize a surface of the form over a rectangular domain. We let and , and let . In this instance, we have , for ,. This surface is graphed in Figure 15.5.5.
We can parametrize the circular boundary of with the vector-valued function , where . We can obtain the interior of by scaling this function by a variable amount, i.e., by multiplying by :, where .
It is important to understand the role of in the above function. When , we get the boundary of , a circle of radius 2. When , we simply get the point , the center of (which can be thought of as a circle with radius of 0). When , we get the circle of radius that is centered at the origin, which is the circle halfway between the boundary and the center. As varies from 0 to 1, we create a series of concentric circles that fill out all of .
Thus far, we have determined the and components of our parametrization of the surface: and . We find the component simply by using :
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Thus ,,, which is graphed in Figure 15.5.7. The way that this graphic was generated highlights how the surface was parametrized. When viewing from above, one can see lines emanating from the origin; they represent different values of as sweeps from an angle of 0 up to . One can also see concentric circles, each corresponding to a different value of .
Examples 15.5.4 and 15.5.6 demonstrate an important principle when parametrizing surfaces given in the form over a region : if one can determine and in terms of and , then follows directly as .
Figure15.5.9.Part (a) shows a graph of the region , and part (b) shows the surface over , as defined in Example 15.5.8
Solution.
We may begin by letting ,, and . This gives only the line on the “upper” side of the triangle. To get all of the region , we can once again scale by a variable factor, .
Still letting ,, we let ,. When , all -values are 0, and we get the portion of the -axis between and . When , we get the upper side of the triangle. When , we get the line , which is the line “halfway up” the triangle, shown in the figure with a dashed line.
Letting , we have ,,. This surface is graphed in Figure 15.5.9.(b). Again, when one looks from above, we can see the scaling effects of : the series of lines that run to the point each represent a different value of .
Another common way to parametrize the surface is to begin with ,. Solving the equation of the line for , we have , leading to using ,. With , we have ,,.
Figure15.5.11.Part (a) shows a graph of the region , and part (b) shows the surface over , as defined in Example 15.5.10
Solution.
While the region in this example is very similar to the region in the previous example, and our method of parametrizing the surface is fundamentally the same, it will feel as though our answer is much different than before.
We begin with letting ,. We may be tempted to let ,, but this is incorrect. When , we obtain the upper line of the triangle as desired. However, when , the -value is 0, which does not lie in the region .
We will describe the general method of proceeding following this example. For now, consider ,. Note that when , we have , the upper line of the boundary of . Also, when , we have , which is the lower boundary of . With , we determine ,,.
Given a surface of the form , one can often determine a parametrization of the surface over a region in a manner similar to determining bounds of integration over a region . Using the techniques of Section 14.1, suppose a region can be described by ,, i.e., the area of can be found using the iterated integral
When parametrizing the surface, we can let ,, and we can let ,. The parametrization of is straightforward, but look closely at how is determined. When ,. When ,.
One can do a similar thing if is bounded by ,, but for the sake of simplicity we leave it to the reader to flesh out those details. The principles outlined above are given in the following Key Idea for reference.
Let a surface be the graph of a function , where the domain of is a closed, bounded region in the -plane. Let be bounded by ,, i.e., the area of can be found using the iterated integral , and let .
The equation can be envisioned to describe an ellipse in the -plane; as the equation lacks a -term, the equation describes a cylinder (recall Definition 11.1.18) that extends without bound parallel to the -axis. This ellipse has a vertical major axis of length 4, a horizontal minor axis of length 2, and is centered at the origin. We can parametrize this ellipse using sines and cosines; our parametrization can begin with
??? ,
where we still need to determine the component.
While the cylinder is satisfied by any value, the problem states that all values are to be between and . Since the value of does not depend at all on the values of or , we can use another variable, , to describe . Our final answer is
Figure15.5.16.The elliptic cone as described in Example 15.5.15
Solution.
One way to parametrize this cone is to recognize that given a value, the cross section of the cone at that value is an ellipse with equation . We can let , for and then parametrize the above ellipses using sines, cosines and .
We can parametrize the component of our surface with and the component with , where . Putting all components together, we have
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When takes on negative values, the radii of the cross-sectional ellipses become “negative,” which can lead to some surprising results. Consider Figure 15.5.17, where the cone is graphed for . Because is negative below the -plane, the radii of the cross-sectional ellipses are negative, and the opposite side of the cone is sketched below the -plane.
Figure15.5.19.An ellipsoid in (a), drawn again in (b) with its domain restricted, as described in Example 15.5.18
Solution.
Recall Key Idea 11.2.25 from Section 11.2, which states that all unit vectors in space have the form for some angles and . If we choose our angles appropriately, this allows us to draw the unit sphere. To get an ellipsoid, we need only scale each component of the sphere appropriately.
The -radius of the given ellipsoid is 5, the -radius is 1 and the -radius is 2. Substituting for and for , we have
,
where we still need to determine the ranges of and .
Note how the and components of have and terms, respectively. This hints at the fact that ellipses are drawn parallel to the -plane as varies, which implies we should have range from to .
One may be tempted to let as well, but note how the component is . We only need to take on values between and once, therefore we can restrict to .
The final parametrization is thus
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In Figure 15.5.19.(b), the ellipsoid is graphed on , to demonstrate how each variable affects the surface.
Parametrization is a powerful way to represent surfaces. One of the advantages of the methods of parametrization described in this section is that the domain of is always a rectangle; that is, the bounds on and are constants. This will make some of our future computations easier to evaluate.
Just as we could parametrize curves in more than one way, there will always be multiple ways to parametrize a surface. Some ways will be more “natural” than others, but these other ways are not incorrect. Because technology is often readily available, it is often a good idea to check one’s work by graphing a parametrization of a surface to check if it indeed represents what it was intended to.
It will become important in the following sections to be able to compute the surface area of a surface given a smooth parametrization ,,. Following the principles given in the integration review at the beginning of this chapter, we can say that
where represents a small amount of surface area. That is, to compute total surface area , add up lots of small amounts of surface area across the entire surface . The key to finding surface area is knowing how to compute . We begin by approximating.
Let be the region of the - plane bounded by , as shown in Figure 15.5.21.(a). Partition into rectangles of width and height , for some . Let be the lower left corner of some rectangle in the partition, and let and be neighboring corners as shown.
The point maps to a point on the surface , and the rectangle with corners , and maps to some region (probably not rectangular) on the surface as shown in Figure 15.5.21.(b), where and . We wish to approximate the surface area of this mapped region.
Let and . These two vectors form a parallelogram, illustrated in Figure 15.5.21.(c), whose area approximates the surface area we seek. In this particular illustration, we can see that parallelogram does not particularly match well the region we wish to approximate, but that is acceptable; by increasing the number of partitions of , and shrink and our approximations will become better.
From Section 11.4 we know the area of this parallelogram is . If we repeat this approximation process for each rectangle in the partition of , we can sum the areas of all the parallelograms to get an approximation of the surface area :
From our previous calculus experience, we expect that taking a limit as will result in the exact surface area. However, the current form of the above double sum makes it difficult to realize what the result of that limit is. The following rewriting of the double summation will be helpful:
(This limit process also demonstrates that and are tangent to the surface at . We don’t need this fact now, but it will be important in the next section.)
In Example 15.5.6, we parametrized the surface as , for ,. To find the surface area using Theorem 15.5.22, we need . We find:
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Thus the surface area is
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There is a lot of tedious work in the above calculations and the final integral is nontrivial. The use of a computer-algebra system is highly recommended.
In Section 15.1, we recalled the arc length differential . In subsequent sections, we used that differential, but in most applications the “” part of the differential canceled out of the integrand (to our benefit, as integrating the square roots of functions is generally difficult). We will find a similar thing happens when we use the surface area differential in the following sections. That is, our main goal is not to be able to compute surface area; rather, surface area is a tool to obtain other quantities that are more important and useful. In our applications, we will use , but most of the time the “” part will cancel out of the integrand, making the subsequent integration easier to compute.
In the following exercises, find the surface area of the given surface . (The associated integrals are computable without the assistance of technology.)
In the following exercises, set up the double integral that finds the surface area of the given surface , then use technology to approximate its value.